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DEF. (Formal infinite series)

A (formal) infinite series is any expression of the form $\sum_{n=m}^\infty a_n$, where $m$ is an integer, and $a_n$ is a real number for any integer $n \geq m$.

DEF. (Convergence of series)

Let $\sum_{n=m}^\infty a_n$ be a formal infinite series. For any integer $N \geq m$ we definite the $N^{th}$ partial sum $S_N$ of this series to be $S_N:=\sum_{n=m}^N a_n$. We say that the infinite series is convergent if the sequence $(S_N)_{N=m}^\infty$ converges to some limit $L$ as $N \to \infty$.

EXERCISE.

Let $\sum_{n=m}^\infty a_n$ be a formal series of real numbers.

Prove that $\sum_{n=m}^\infty a_n$ converges iff for every real number $\varepsilon >0$ there exists an integer $N \geq m$ such that $|\sum_{n=p}^q a_n| \leq \varepsilon$ for all $p, q \geq N$.

My attempt (rightward implication):

since $\sum_{n=m}^\infty a_n$ converges, by definition of convergent infinite sequence we have that the sequence $(S_N)_{N=m}^\infty$ converges to some limit $L$ as $N \to \infty$; this implies that $(S_N)_{N=m}^\infty$ is a Cauchy sequence, so that if we fix $\varepsilon >0 $ there exists $N' \geq m$ s.t. $|S_n - S_{n'}|\leq \varepsilon$ $\forall n, n' \geq N'$ by definition of Cauchy sequence. Therefore if we take $q \geq p \geq N'$ arbitrarily, we have $|S_q-S_p|= |\sum_{n=m}^q a_n - \sum_{n=m}^p a_n|=|\sum_{n=p+1}^q a_n| \leq \varepsilon.$ The thesis now follows form the fact that $\varepsilon >0$ and $p,q \geq N$ were chosen arbitrarily.

I think this proof would work if only I could replace $p+1$ with $p$ as the first index of the last summation by I haven't been able to do so, so I would appreciate any hint about how to accomplish this.

Best regards,

lorenzo.

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    $\begingroup$ It wouldn't be a problem at all, because $p>N'$ implies $p+1>N$. $\endgroup$ – Chee Han Jan 22 '16 at 16:31
  • $\begingroup$ @CheeHan : I don't get what you are suggesting; could you explain? $\endgroup$ – lorenzo Jan 22 '16 at 16:45
  • $\begingroup$ @lorenzo, I wrote out the work in an answer. More or less, the same as Chee Han's comment. $\endgroup$ – Andres Mejia Jan 23 '16 at 0:11
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For the rightward implication, you're right that there's something weird about that last step. The lower index being $p+1$ is only accurate when $q>p$. Since we need to account for $q \geq p$, we then have a second (much more trivial) case where $q=p$.

The case where $q=p$ is in fact so trivial that some instructors might not even care if you actually evaluate the steps. But it's nevertheless essential to acknowledge the difference between the two cases in your proof. Because this lies at the root of your confusion. You are able to index the bottom index $p+1$ after taking the difference ONLY IF $q>p$.

As a side note, you could equivalently define a Cauchy Sequence by using arbitrary $q>p \geq N'$, just because the case where $q=p$ we get for free! But most textbooks like the convention "$q \geq p \geq N'$" or "$q \geq N'$ and $p \geq N'$."

EDIT: DISREGARD the previous two paragraphs (but not their insights about the nature of sequences and series). Sorry for the misunderstanding. Here's the problem with your strategy: You're giving your partial $p$ and $q$ sums too much wiggle room. You need to size them down using the "$\epsilon/2$ trick" and then plop the extra term $a_p$ in with the remaining $\epsilon/2$. Note if you replace $\epsilon$ with $\epsilon/2$, you can get your $N$ to squeeze $\sum^q_{n=p+1} a_n$ inside of $\epsilon/2$ to get $q \geq p \geq N \implies \left| \sum^q_{n=p+1} a_n \right| \leq \epsilon/2$ (using the steps you posted). For the next step, note series convergence implies sequence convergence to $0$ (using the contrapositive statement of The Divergence Test). Hence there is some $M$ such that $|a_p-0|=|a_p| \leq \epsilon/2$. Take the maximum of $N$ and $M$ and that'll do it. I'm leaving it to you to connect the dots, but the strategy is outlined.

Now here's a big hint for the converse (the leftward implication): Have you heard of the Cauchy Convergence Theorem? You're gonna need it!

EDIT 2: If you're struggling with the converse, I can actually go through the steps myself and give you another hint.

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  • $\begingroup$ Thank you for your answer but even taking the fact that $q$ should be strictly greater than $p$ into account doesn't change the fact that to finish the proof I should get $p$ in the first index of the last summation, and not $p+1$ (see the thesis of the theorem). What puzzles me is how to accomplish this last step I've just said. $\endgroup$ – lorenzo Jan 22 '16 at 20:23
  • $\begingroup$ OOOHHHH, I see now. I know what to do. Stay tuned for an edit. $\endgroup$ – Logician6 Jan 22 '16 at 20:35
  • $\begingroup$ Everything is edited. Hope that helps. $\endgroup$ – Logician6 Jan 22 '16 at 20:53
  • $\begingroup$ Thank you again for your answer, but in the text from which this exercise was taken from, the fact that $\sum_{n=m}^\infty a_n$ is convergent implies $lim_{n\to\infty} a_n = 0$ is stated as a corollary of this theorem, so it can't be used in this proof. $\endgroup$ – lorenzo Jan 22 '16 at 22:20
  • $\begingroup$ REALLY??? I see how that is corollary, but you can prove the divergence test (or rather its contrapositive consequence) in a much simpler way than just proving this more general exercise. Anyway, I think I've got another way that doesn't involve $\epsilon/2$. Hold up. I'll write a new answer. $\endgroup$ – Logician6 Jan 22 '16 at 22:49
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Rightward Implication: Your first few steps involving your use of the Cauchy Convergence Theorem are again correct. But what I think is the missing link is induction on $p$. This really isn't much beyond your scratchwork since your scratchwork ALREADY DOES the inductive step, which interestingly enough doesn't use the inductive hypothesis like induction generally does. But that logically doesn't matter. If we've proven a statement $B$ all by itself, we've thereby proven $A \implies B$, for any statement $A$.

ALL we need now is the base case, which is $p=N$. But that just follows directly from our hypothesis that $\sum^\infty_{n=m} a_n$ converges in the first place.

Leftward Implication: The only thing special that you use is again the Cauchy Convergence Theorem (the part of it that states Cauchy Sequence implies Converging Sequence) if I have it all worked out correctly. All you have to do is show that a series that such that \begin{align*} \forall (\epsilon>0)\exists (N\geq m) \forall(p \geq N)\forall (q \geq p)\left(\left| \sum^q_{n=p} a_n \right| \leq \epsilon \right), \end{align*} is Cauchy, thereby making it convergent, which shouldn't be too bad.

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This is a very small problem. As a rule of thumb, all that matters is what happense eventually, or as $N$ becomes sufficiently large.

Let $q \geq p \geq N'$. Then $p+1>N^{\prime}$.Thus, we have that $|S_q-S_{p+1}|= |\sum_{n=m}^q a_n - \sum_{n=m}^{p+1} a_n|=|\sum_{n=p}^q a_n| \leq \varepsilon.$ Hypothesis follows.

Otherwise, I agree. The forward direction looks good.

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